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Full text of "Text Book Of Mechanical Engineering"

Strength of Suspension Link.

405

Example 10.—Find the thickness of a short, hollow, cast-iron
column of 18" outside diameter, to sustain a live load of 80 tons, plus
a dead load of 100 tons. (Eng\ Exam. 1888.)

Equivalent dead load = 100 + (2 x 80) = 260 tons
260 = 4 a       and a = 65 sq. ins.

But<2=7rR2-7r r2=65         or   — (8i-^2)=6s
andr= V 60*3 =7*8              .'.    /=9 -7*8 = 1 "z ins.

Shear Stress-Action rarely occurs alone, but pins and rivets
are thus calculated :            W = /s #.    *

Strength of a Suspension Link (see Fig. 355).—The
strength of one thin link in tension, at a and c\ the shear

strength of the pin d-} the strength at b] and the bearing stress
on projected area of e, should each equal half the load :
(i)           (2)                 (3)*      (4)         (5)

(w -

Let /t-i, /b=4,   and/,-}.
By(4>and(5)
By (2) and (4)

By (3) and (4)       |x^~i*/

By actual tests      ^           =/

and* the thick link must be 2 / in thickness.

and

'66^

i Kbt      and w=:

and / =  '20 &
«. '66^

p. 415.

(See Appendix IV., #. 952.)